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If $A,B$ non empty, upper bounded sets and $A+B=\{a+b\mid a\in A, b\in B\}$, how can I prove that $\sup(A+B)=\sup A+\sup B$?

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Show that $\sup(A+B)$ is less than or equal to $\sup(A)+\sup(B)$ by showing that the latter is an upper bound for $A+B$. Then show that $\sup(A)+\sup(B)$ is less than or equal to $\sup(A+B)$ by showing that $\sup(A+B)$ is an upper bound for $A+\sup(B)$ and that $\sup(A+\sup(B)) = \sup(A)+\sup(B)$.

Set $x=\sup A$, $y=\sup B$. Given $a\in A, b\in B$ we have $a\le x, b\le y$ so $a+b\le x+y$ so it’s an upper bound. Now take $\varepsilon > 0$ and find $a,b$ such that $a>x-\varepsilon/2, b>y-\varepsilon/2$ and you have $a+b > x+y – \varepsilon$. That suffices (since it means that every potential “smaller upper bound” $x+y-\varepsilon$ is not really an upper bound).

(question 2. http://webcache.googleusercontent.com/search?q=cache:DohoRC3-bU8J:www.maths.usyd.edu.au/u/UG/IM/MATH2962/r/PDF/tut01s.pdf)

The idea is to prove two inequalities:

sup(A+B) ≥ supA+ supB and sup(A + B) ≤ supA + supB.

$\Large{Prove \; \sup A + \sup B \le \sup(A + B)}$:

By definition of A + B and sup(A + B), for all a ∈ A and b ∈ B,

$\color{seagreen}{a + b} – b \quad \color{seagreen}{\le \quad \sup (A + B)} – b$

Hence if we fix $b ∈ B$, then $\color{seagreen}{\sup (A + B)} – b$ is an upper bound for $\color{seagreen}{A + B} – B = A$.

And so by definition of $\sup A$, for every $b ∈ B$, $\sup A ≤ \sup (A+ B) − b $.

Rearrange: $\color{magenta}{b} ≤ \sup(A +B) − \sup A$ for all $b ∈ B$.

Ergo, $\qquad \qquad \sup(A +B) − \sup A$ is an upper bound for any $\color{magenta}{b}$.

So again by the definition of a supremum: $\color{magenta}{\sup B} ≤ \sup(A + B) − \sup A \qquad \iff

\sup A + \sup B ≤ \sup(A + B).$

$\Large{Prove \; \sup A + \sup B \ge \sup(A + B)}$:

Since sup A is an upper bound for A, $a ≤ \sup A$ for all a ∈ A.

Similarly, $b ≤ \sup B$ for all b ∈ B. Hence $a + b ≤ \sup A + \sup B$ for all x ∈ A

and y ∈ B.

Ergo $\sup A + \sup B$ is an upper bound for A + B.

Hence by definition of a supremum, $\sup A + \sup B \ge \sup(A + B)$.

Another proof for $\sup(A + B) \ge \sup A + \sup B$.

$\Large{\sup A + \sup B \; is \; finite.}$

Posit $e > 0.$ Then there

exists $a \in A$ and $b \in B$ such that

$x > \sup A − \frac{e}{2}$ and $y > \sup B − \frac{e}{2}$.

Then $a + b \in A + B$.

Ergo $\color{seagreen}{sup(A + B)} \ge a + b \color{seagreen}{> \sup A + \sup B – e}$.

$\implies \color{seagreen

}{ \sup(A + B) > \sup A + \sup B – e }$.

Since $e > 0$ is arbitrary, $\sup(A + B) \ge \sup A + \sup B$

$\Large{\sup A + \sup B \; is \; infinite.}$

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